If the solubility product of AB_2is 3.20 10^-11M^3, then the solubility of AB_2 in pure water is 10^-4 molL^-1. [Assuming that neither kind of ion reacts with water]

Question

Accepted Answer

Correct answer: 2. (2) AB_2 ⇌A_s^2+ ( aq )+2B_2s^- ( aq ) K_ sp=4s^3=3.2 10^-11 ⇒s^3=8 10^-12 s=2 10^-4

$If the solubility product of AB_{2} is 3.20 \times 10^{- 11} M^{3}, then the solubility of A B_{2} in pure water is \times 10^{- 4} molL^{- 1} . [Assuming that neither kind of ion reacts with water]$

Solution

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